Question

Solve for x in the proportion

Answer 1

\[\frac{ 9 }{ 27x }=\frac{ x }{ x+3 }\]

Answer 2

A.x = 2 and x = 0

B. x ≈ −2.48 and x ≈ 2.48

C. x ≈ −0.85 and x ≈ 1.18

D. x ≈ −6 and x ≈ 6.16

Answer 3

multiply both sides by (x+3)27*x

Answer 4

First, decide \(x \ne 0\;and\;x \ne -3\). Then you can proceed.

Answer 5

its assumed x is not 0 and x is not -3

Answer 6

or use cross multiply to get rid off the fractions

Answer 7

we only need to note it when we divide away that information

Answer 8

if we want equality to hold...

Answer 9

so the answer?

Answer 10

multiply both sides by 27x(x+3)
and tell us what you have

Answer 11

idk? im redoing a test from freshman year its multiple choice i just need an answer

Answer 12

:)

Answer 13

we are not here to give answers, or do your test....
27x^2=(x+3)9

Answer 14

k

Answer 15

this is what we have after we multiply both sides by 27x(x+3), do you see why?

Answer 16

ya

Answer 17

\[27x^2=9(x+3)\]
expand the right hand side and what do we get?

Answer 18

idfk bye

Answer 19

Uhh should be C lol

Answer 20

@jefftheloveableguy read the rules, we are not here to take someones test....

Answer 21

Disagree with the "assumed" part. It is mandatory.

Answer 22

it is assumed by the equation itself

Answer 23

1/x is assumed that x is not 0

Answer 24

if we divide away the x we need to note that....

Answer 25

the domain of 1/x is not R, it is R\{0} so its assumed x is not 0.
understand @tkhunny ?

Answer 26

x/x has the same domain, but if we write it as x/x = 1, we need to note that the domain of f(x) = 1 = x/x if and only if x is not equal to 0

Answer 27

Still disagree. Equations don't know anything. Humans do. :-)

In my book, nothing is assumed until you state explicitly that it is so. Lacking such explicit statements, I would mark it wrong on an exam.

Answer 28

ok .... if we difine a function we define a domain, you cant have a function without a domain
the domain of 1/x is all real number that are not 0
so we do not need to note it outside of that...
it is not until we get rid of that information by dividing away something that we need to make that note.

Answer 29

believe me or not....my note to you was simply about you saying we need to define the domain. before we move on. this is not the point, the point is to make sure the domains match when we are done....or we don't have equality

Answer 30

Not a matter of belief. I prefer to observe the Domain from the beginning, rather than the riskier back-end check. Clearly (and not just in this conversation), not everyone agrees with me. Most students find my insistence quite annoying, until later in the course when they observe the value of the stricter exercise.

Answer 31

Good discussion, thank you.

Answer 32

thanks both for your help. :)

Answer 33

1: first off you are assuming that this is defined on the real line, it could be the natural numbers and then we don't even need to worry. so you are making assumptions about it being a real function in general, as im sure at some point in the book it says assume the functions are real

2:say I ask the question
evaluate f(3) if f(x) - 3x+3
and then you say , well first we need to note this is defined on the real numbers, and thus 3 is defined

the main point
\[f:R-\{0\}\rightarrow R,f(x)=\frac{x}{x} \]
there is no point in stating the domain at first because of course x cant be 0
but if we conclude
\[f(x) = 1\]we must say something about the domain, because f(x) is not defined at 0, this is easy to see when we have f(x) = x/x, but not so easy to see(impossible unless guessed) with f(x) = 1, so we note...f(x) = 1, iff x is not 0

Answer 34

sorry had to finish my thought:)

Answer 35

Okay everyone is entitled to their own opinion. lease no arguments on my question. have a good night thank u both :)

Answer 36

:-)

Answer 37

not arguing:)